Aromatic carbonyl compound having molecular formula `C_(8)H_(8)O` react with `NH_(2)OH` how many oximes can be formed

Options

A. 8
B. 10
C. 12
D. 6

Answer

To determine how many oximes can be formed from an aromatic carbonyl compound with the molecular formula C8H8O when it reacts with hydroxylamine (NH2OH), we can follow these steps:

### Step 1: Identify the structure of the carbonyl compound
The molecular formula C8H8O suggests that the compound is likely an aromatic carbonyl compound. The double bond equivalent (DBE) can be calculated to confirm the presence of a benzene ring.

**Calculation of DBE:**
DBE = C - (H/2) + (N/2) + 1
- C = 8
- H = 8
- N = 0
- Halogens = 0

DBE = 8 - (8/2) + (0/2) + 1 = 8 - 4 + 1 = 5

Since the DBE is greater than 4, it indicates that there is a benzene ring present in the compound.

### Step 2: Possible structures of the carbonyl compound
Given the molecular formula C8H8O, the following aromatic carbonyl compounds can be considered:
1. Acetophenone (C6H5COCH3)
2. 2-Methylbenzaldehyde (C6H5CHO)
3. 3-Methylbenzaldehyde (C6H5CHO)
4. 4-Methylbenzaldehyde (C6H5CHO)
5. Benzylacetaldehyde (C6H5CH2CHO)

### Step 3: Reaction with hydroxylamine
When each of these carbonyl compounds reacts with hydroxylamine, they form oximes. The general reaction can be represented as:
\[ \text{R-C(=O)-R'} + \text{NH}_2\text{OH} \rightarrow \text{R-C(=N-OH)-R'} + \text{H}_2\text{O} \]

### Step 4: Determine the number of isomers for each compound
1. **Acetophenone**: Forms 2 isomers (E and Z).
2. **2-Methylbenzaldehyde**: Forms 2 isomers (E and Z).
3. **3-Methylbenzaldehyde**: Forms 2 isomers (E and Z).
4. **4-Methylbenzaldehyde**: Forms 2 isomers (E and Z).
5. **Benzylacetaldehyde**: Forms 2 isomers (E and Z).

### Step 5: Total number of oximes formed
Each of the 5 compounds can form 2 isomers upon reaction with hydroxylamine:
Total oximes = 5 compounds × 2 isomers/compound = 10 oximes.

### Final Answer
Thus, the total number of oximes that can be formed is **10**.

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